A) \[{{10}^{-5~~}}\]
B) \[1.11\text{ }\times \text{ }{{10}^{-6}}\]
C) \[1.11\text{ }\times \text{ 1}{{0}^{-4~~}}\]
D) 0.01
Correct Answer: B
Solution :
\[{{\Lambda }_{m}}(HC)={{\Lambda }_{m}}(HCl)+{{\Lambda }_{m}}(NaC)-\Lambda m(NaCl)\] |
\[=426+83-126=383\,\,{{\Omega }^{-\,1}}\,c{{m}^{2}}\,mo{{l}^{-\,1}}\] |
Molar conductivity of HC |
\[{{\Lambda }_{m}}(HC)=\frac{k}{C}\]\[=\frac{3.83\times {{10}^{-5}}{{\Omega }^{-1}}c{{m}^{-1}}}{0.001}\times 1000\]\[=3.83\,{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}\] |
\[\alpha =\frac{{{\Lambda }_{m}}(HC)}{\Lambda _{m}^{\infty }(HC)}=\frac{38.3}{383}=0.1\] |
\[{{K}_{\alpha }}=\frac{C{{\alpha }^{2}}}{1-\alpha }=\frac{{{10}^{-\,3}}\times {{0.1}^{2}}}{(1-0.1)}=1.11\times {{10}^{-\,5}}\] |
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